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Coordinate Geometry – The equation of a circle

📘 Notes
Equation of a Circle — Notes 1. Standard form

The circle with centre \((a,b)\) and radius \(r\) has equation \( (x-a)^2 + (y-b)^2 = r^2. \) Centre = \((a,b)\). Radius = \(r\).

Example: \((x-2)^2 + (y+3)^2 = 16\) → centre \((2,-3)\), radius \(4\).

2. General form

A circle can be written as \( x^2 + y^2 + 2gx + 2fy + c = 0. \) Then the centre is \((-g,-f)\) and the radius is \(\sqrt{g^2+f^2-c}\).

Converting to standard form (complete the square)

Group \(x\)-terms and \(y\)-terms, complete squares and move the constant:

Given: \( x^2 + y^2 - 4x + 6y - 12 = 0 \) Group: \( (x^2 - 4x) + (y^2 + 6y) = 12 \) Complete: \( (x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9 \) Result: \( (x-2)^2 + (y+3)^2 = 25 \) → centre \((2,-3)\), radius \(5\).

3. Circle with given centre and radius

Directly substitute into standard form: centre \((a,b)\), radius \(r\):

\((x-a)^2 + (y-b)^2 = r^2\).

4. Circle with diameter endpoints

If \(A(x_1,y_1)\) and \(B(x_2,y_2)\) are ends of a diameter, one convenient form is \( (x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0, \) or equivalently the circle with centre at the midpoint of \(AB\) and radius half the distance \(AB\).

5. Finding centre and radius — summary steps

  1. Write the equation as a polynomial: \(x^2 + y^2 + 2gx + 2fy + c = 0\) (if necessary divide through by a constant).
  2. Complete the square for \(x\) and \(y\): add and subtract the required constants.
  3. Rearrange into \((x-a)^2 + (y-b)^2 = r^2\).
  4. Read off the centre \((a,b)\) and radius \(r\).

6. Special cases & cautions

  • If \(g^2 + f^2 - c < 0\) then there is no real circle (radius would be imaginary).
  • If radius \(= 0\) the circle reduces to a single point (the centre).
  • If the equation has coefficients multiplied (e.g. \(2x^2+2y^2+\dots=0\)), divide by that factor first.

7. Quick reference table

FormEquationCentreRadius
Standard \((x-a)^2+(y-b)^2=r^2\) \((a,b)\) \(r\)
General \(x^2+y^2+2gx+2fy+c=0\) \((-g,-f)\) \(\sqrt{g^2+f^2-c}\)
Diameter AB \((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\) midpoint of \(A,B\) half distance \(AB\)

8. Examples (worked)

Example 1: \(x^2+y^2-6x+8y+10=0\)

\( (x^2-6x)+(y^2+8y)=-10 \) \( (x-3)^2+(y+4)^2 = 15 \) Centre \((3,-4)\), radius \(\sqrt{15}\).

Example 2: \(x^2+y^2-x-10y+5=0\)

\( (x^2-x)+(y^2-10y)=-5 \) \( (x-\tfrac{1}{2})^2+(y-5)^2 = \tfrac{81}{4} \) Centre \((\tfrac{1}{2},5)\), radius \(\tfrac{9}{2}\).

Example 3 (scale present): \(2x^2+2y^2-8x+2y+2=0\)

Divide by 2: \( x^2+y^2-4x+y+1=0 \) Complete squares: \( (x-2)^2+(y+\tfrac{1}{2})^2=\tfrac{13}{4} \) Centre \((2,-\tfrac{1}{2})\), radius \(\tfrac{\sqrt{13}}{2}\).